# An Inequality Where One Term Is More Equal Than Others

### Proof 1

By the Cauchy-Schwarz inequality,

$\displaystyle\left(a_1 +\sum_{k=2}^na_k\right)\left(\frac{(n-1)^2}{a_1}+\sum_{k=2}^n\frac{1}{a_k}\right)\ge 4(n-1)^2.$

Then

\displaystyle\begin{align} \left(\sum_{k=1}^na_k\right)&\left(\sum_{k=1}^n\frac{1}{a_k}\right)\ge 4(n-1)^2\\ &\ge 4(n-1)^2-\frac{n(n-2)\displaystyle\sum_{k=1}^na_k}{a_1}\\ &\ge 4(n-1)^2-\frac{2n(n-2)a_1}{a_1}\\ &= 4(n-1)^2 - 2n(n-2)\\ &= n^2 + (n-2)^2. \end{align}

### Proof 2

Let $a=\displaystyle\sum_{k=2}^na_k\,$ $x=a_1\,$ and $y=\displaystyle\sum_{k=2}^n\frac{1}{a_k}.\,$ Clearly, $x\ge a\gt 0\,$ and, by the AM-HM inequality, $y\ge\displaystyle\frac{(n-1)^2}{a}.$

We need to prove that $\displaystyle (x+a)\left(\frac{1}{x}+y\right)\ge 2[(n-1)^2+1]\,$ which is equivalent to $(x+a)(xy+1)\ge 2x[(n-1)^2+1].\,$ But $\displaystyle xy+1\ge \frac{(n-1)^2x+a}{a},\,$ hence, suffice it to prove that

$(x+a)[(n-1)^2x+a]\ge 2ax[(n-1)^2+1].$

This can be rewritten as

$(n-1)^2x^2-[(n-1)^2+1]ax+a^2\ge 0,$

or

$(x-a)[(n-1)^2x-a]\ge 0,$

which is true because $x\ge a\ge\displaystyle\frac{a}{(n-1)^2}.$

Note that equality holds iff $a_2=\ldots=a_n=t\,$ and $a_1=(n-1)t,\,$ $t\gt 0.$

### Acknowledgment

Leo Giugiuc has kindly pointed to me the discussion at the mathematical inequalities facebook group of the problem by Nguyen Viet Hung, where it was commented on by Sarah El (Proof 1) and Leo Giugiuc (Proof 2).